22.4 litres of a gas weighs 70 g at S.T.P. Calculate the weight of the gas if it occupies a volume of 20 litres at $27^{\circ} C$ and 700 mm Hg of pressure.

22.4

litres of a gas weighs

70

g at S.T.P. Calculate the weight (gm) of the gas if it occupies a volume of

20

litres at

27^{o}

C and

700

mm Hg of pressure.

Q. 16 g of oxygen and 3 g of hydrogen are mixed and kept at 760 mm pressure and 0$$. The total volume occupied by the mixture will be nearly
(A) 22.4 litre
(B) 33.6 litres
(C) 448 litres
(D) 44800 ml

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The total weight of a gas mixture which contains 11.2 litres of CO2, 5.6 litres of O2, 22.4 litres of N2, 44.8 litres of He at 0oC and 760 mm of Hg pressure is:

The correct option is C 66 gSince given conditions are STP, we can say that there are 0.5 moles of CO2 , 0.25 moles of O2 , 1 mole of N2 and 2 moles of He.W=W1+W2+W3+W4W =22+8+28+8=66 g

The total weight of a gas mixture which contains $11.2$ litres of $C O_{2}$ , $5.6$ litres of $O_{2}$ , $22.4$ litres of $N_{2}$ , $44.8$ litres of $H e$ at $0^{o} C$ and $760$ $m m$ of $H g$ pressure is:

The correct option is C $66$ $g$
Since given conditions are $S T P$ , we can say that there are $0.5$ moles of $C O_{2}$ , $0.25$ moles of $O_{2}$ , $1$ mole of $N_{2}$ and $2$ moles of $H e$ .
$W = W_{1} + W_{2} + W_{3} + W_{4}$
$W$ = $22 + 8 + 28 + 8 = 66$ $g$